PREPARATORY EXAMINATIONS

VOORBEREIDENDE EKSAMEN

2018

MARKING GUIDELINES /

NASIENRIGLYNE

MATHEMATICS / WISKUNDE

(PAPER 2 / VRAESTEL 2) (10612)

15 pages / bladsye

10612

2

GAUTENG DEPARTMENT OF EDUCATION

GAUTENGSE DEPARTEMENT VAN ONDERWYS

PROVINCIAL EXAMINATION / PROVINSIALE EKSAMEN

MATHEMATICS / WISKUNDE

(PAPER 2 / VRAESTEL 2)

MARKING GUIDELINES /

NASIENRIGLYNE

QUESTION / VRAAG 1

1.1 24a 24a (2)

1.2 100 100 (1)

1.3 6050 x answer (1)

1.4.1

2,39

100

553445283582561524

x

Answer only 3/3

numerator

2,39

(3)

1.4.2 100 – 34

= 66 34

66 (2)

[9]

QUESTION / VRAAG 2

2.1 47,6 answer (1)

2.2 5,14x One standard deviation above the mean:

97,20

47,65,41

Therefore a student needed to work for 21 hours

5,14x

20,97 21 hours

(3)

2.3 xy 42,13138,454 a = 38,454

b = 42,131

equation /

vergelyking (3)

2.4 Payment when 5,11x / Betaling wanneer 5,11x

marks 2 / 2 965,731 R71,1965 R

calculator theusingby 5,1142,13138,454

42,13138,454

OF OR /xy

subst x = 11,5 / vervang x = 11.5

71,1965R

(2)

2.5 2700;23

The student could have done some of the work incorrectly and

needed to redo work without any payment

Any logical reason can be accepted

2700;23

reason

(2)

[11]

10612

3

QUESTION / VRAAG 3

3.1.1

1

)2(2

)3(1AD

m

3

14

1

41

34

)4(1

]BC||AD[1BCAD

r

r

r

r

mm

OR / OF

3

14

34

4

22

31

]BC||AD[BCAD

r

r

r

mm

1AD m

34

)4(

r

14 r value of r / waarde van r

1AD m

34

)4(

r

14 r value of r /

waarde van r (4)

3.1.2 Trapezium / Trapesium answer / antwoord

(1)

3.1.3

)1;0(P

2

13;

2

22P

OR / OF

1;0P

1

211

xy

xy

x = 0

y = –1 (2)

3.1.4

ADBP

1

11

1

30

)4(1

PBAD

PB

mm

m

1PB m

product = –1

(2)

y

x

D(2 ; 1)

x

C(4 ; r)

B(3 ; –4)

A(–2 ; –3)

P

10612

4

3.1.5 AB is a diameter / middellyn [ º90BPA ]

2

13or

4

26

2

26

26AB

2

r

r

2

13

2

7

2

1

2

7;

2

1

2

34;

2

23ABofmidpoint

22

yx

OR / OF

2

7;

2

1

2

34;

2

23ABofmidpoint

2

13

2

13

4

25

4

1

4

1

4

25

2

71

2

10/or3

2

72

2

122

2

22

2

rofr

2

13

2

7

2

122

yx

2

26r

2

132 r

2

1

2

7

equation / vergelyking

2

1

2

7

substitution into distance formula /

vervang in afstandformule

2

132 r

equation /

vergelyking

(5)

3.1.6

3ofvalueMaximum

52

1cos of maximum the of any valueFor

cos52

cos4cos2

cos4cos2cos2cos

2cos4cos2

2cos4cos2

2

2

2

222

2222

22

22

r

r

r

yx

yyxx

yxyx

22

cos2cos yx

22 cos4cos2

22 cos52r

1cos ofmax 2

3r

(5)

middelpunt van AB

middelpunt van AB

10612

5

3.2

3.2.1

0153

5

1

5

3

5

10153

5

926355

)3(5

32or / )2(

5

31

5

3

23

12PQ

yx

xy

cyx

cxy

cofxy

m

gradient of PQ / gradiënt van PQ

sub P, Q and / en m

equation / vergelyking (correct form)

(3)

3.2.2

º04,149

º96,30º180OSP

5

3

OSPtan PQ

m

7373

)2(31or/)2(31

3

º57,71tan

º57,71

/]Δ of ext [º47,77º04,149STP

PR

xyxy

cofxy

m

5

3OSPtan

OSP = º04,149

STP =71,57º

PRm = 3

sub P(–2;1) and / en m equation /vergelyking

(correct form) (6)

[28]

[buite van ∆]

10612

6

QUESTION / VRAAG 4

4.1

)3;2(C

25)3()2(

9412)3()2(

1246

22

22

22

yx

yx

xyyx

Answer only 3 / 3

22 )3()2( yx

)3;2(C

(3)

4.2 5r 5r (1)

4.3 DC2 = (2 + 1)2 + (–3 + 2)2 = 9 + 1

=10

BC2 = 25 (radius) DB2 = BC2 – DC2 (Pyth)

= 25 – 10 = 15

DB = 15

AB = 2 15 or / of 7,75

sub into distance formula / vervang in

afstandformule DC2 = 10

sub into / in Pythagoras

DB = 15

AB = 2 15 or / of

7,75 (5)

4.4

25,12/65

101522

1

)DC)(AB(2

1Area

area formula /

oppervlakte formule

substitution / vervanging

(ca from / van 4.2)

answer / antwoord

(3)

[12]

y

x

C

B

D(–1 ; –2)

A

O

10612

7

QUESTION / VRAAG 5

5.1

yxyx.

yx.yx.

sinsin360coscos

180sincos90sinsin

)tan(

)cos(

)sin(

sin.sincos.cos

sin.coscos.sin

sin).sin(cos.cos

)sin.(coscos.sin

yx

yx

yx

yxyx

yxyx

yxyx

yxyx

ycos

ysin

ycos

xsin

)cos(

)sin(

yx

yx

)tan( yx

(6)

5.2.1

RHS

sinAsinBcosAcosB

B)sin(A)sin(B)cosAcos(

B))(cos(A

B)cos(A

B))(cos(A

subst BB

(2)

5.2.2(a)

22

222

1 ax

ryx

aa ;1T 2

Pythagoras

aa ;1T 2 (2)

Answer only 2/2

5.2.2 (b) aa ;1R 2 aa ;1R 2

(2)

5.2.2

(c)

1

1

)(11

QsinPsinQcosPcos)QPcos(

22

22

aa

aaaa

22 11 aa

aa ( ) 1

(2)

5.2.2(d)

180Q;90P;360QP

1360cos

answer (1)

P

1

T

10612

8

5.3.1 11 d or / of ]1;1[ answer / antwoord (2)

5.3.2

0)3cos2)(2cos3(

06cos5cos6

cos56cos6

6

5

cos

1cos

2

2

Zkk

;360.81,131

oplossingeen solution/g no2

3cosor

3

2cos

OR / OF

ZkkZkk

kk

;360.19,228;360.81,131

360.19,48180or360.19,48180

06cos52cos6 factors / faktore

both solutions

choosing 3

2cos

360.81,131 k

360.81,131 k

OR / OF

360.19,228 k

If Zk is omitted then

subtract one mark / As Zk weggelaat is, trek

een punt af (6)

[23]

10612

9

QUESTION / VRAAG 6

6.1 45x 45x (has to be an equation x =..)

(1)

6.2

)(

1

2tan

1

2sin

2cos

cos.sin2

)sin21(

cos.sin2

)sin21(sin

cos.sin2

sin2sin)(

2

2

2

2

3

xf

x

x

x

xx

x

xx

xx

xx

xxxh

)2

sin21(sin xx

x2cos

x2sin

)(

1

xf

(4)

6.3

x

x

xxp

2cos1

)902sin(1

)45(2sin1)(

Answer only 3/3

)45(2sin1 x

)902sin( x

x2cos1

(3)

6.4

0)2sin1.(2tan

0)2sin1.(2tan

xx

xx

4590 x or )45;90[

0)2sin1.(2tan xx

x90

45x (3)

[11]

f

g

-135 -90 -45 45

90

•

•

10612

10

QUESTION / VRAAG 7

7.1

)32(

32

2

322

30cos.2

DE2EG.EDcosGEDEGGD

2

22

22

22

222

p

pp

pp

pppp

correct substitution in cos-rule / korrekte vervanging in cos-reël

2

3

simplification /

vereenvoudiging

22 32 pp

(3)

7.2 In CDGΔ

336

3).32(

3.)32(

60tan.GDCD

DGCtanGD

CD

2

2

p

p

p

DGCtanGD

CD

)32(2 p

3

(3)

[6]

10612

11

QUESTION / VRAAG 8

8.1 EABC

EBCD

EABC

EBCD (2)

8.2.1 ] /theoremchordtan[º40B2 koord en raaklyn tussen SR (2)

8.2.2 BC||AE;./]BC||AE;alt[º60B1

es binneverw S R

(2)

8.2.3 ]/quad cyclicaofopp[º120D s kvh van teenoorst. e SR

(2)

8.2.4

] /Δaofofsum[º30E

]/sidesequalopp[CE

s

1

es

11

Δvan/Δvansom

syegelyketeenoor

ee

S S

R (3)

8.3 AD = 20

90CDA [line from centre to midpoint of chord / midpt. ʘ ; midpt. koord OF

lyn van midpt ʘ na midpt van koord]

03,36C

588,0Csin

34

20Csin

Accept 36º

AD = 20

R

34

20Csin

36C (4)

[15]

10612

12

QUESTION / VRAAG 9

9.1

)//Δaofs'ofsum(290B

)/radiiequalopposites'(CB

OBCIn

2

32

Δvan/Δvansomx

radii gelyketeenoor

ee

e

S / R

x290B2

(2)

9.2

x

syegelyketeenoor

gegeemaar

vanbuite

omtreksmidpts

x

E

)/sidesequalopps'(EC

) /(givenAEACBAbut/

)Δ/Δaof(ext.ECA

)2

ncecircumfereat2centreat(2A

e

1

13

3

SR

S

S

xE (5)

9.3

x

Δvan/Δvanm

xxx

syegelyketeenoor

ee

2

21

e

3221

C

)so

/Δa of s'of(sum)2902902(180CB

)

/sidesequalopps'(CCBB

S

S

S (3)

9.4

)

90 ( 'ED

circle) semi ain ofconverse/

90subtends(lineAEDcircleof diameter a isED

90A

290

)

/ral quadrilate cyclicaof(ext.CA

1

1

sirkelhalweinvanomgekeerde

ofonderspanlynAEDsirkelvanmiddellynnis

xxx

hoekkoordeviervanbuite

S R

90A1

R

(4)

[14]

10612

13

QUESTION / VRAAG 10

10.1 Construction: In ADE, draw height h relative to base AD and height k relative to base AE. Join BE and DC to create BDE and

CED. /

Konstruksie: In ADE, trek hoogte h relatief tot basis AD en die

hoogte k relatief tot basis AE. Verbind BE en DC om BDE en

CED te vorm.

Proof: / Bewys:

EC

AE

DB

AD

ΔCEDArea

ΔADEArea

ΔBEDArea

ΔADEArea

]

height /samebase,[same / ofAreaBEDofAreabut

EC

AE

.CE2

1

.AE2

1

ΔCEDArea

ΔADEArea

DB

AD

.B2

1

.AD2

1

ΔBEDArea

ΔADEArea

hoogte dies. basis; dies.

CEDvan

k

k

hD

h

construction / konstruksie

S

S

SR

S

(6)

A

D

B C

E

h k

10612

14

10.2

10.2

XY]AB

/XY||AB ; theoremprop.

of side one to||line[DY

DB

DX

DA :DXYIn

|| ; stellingewer.

ofΔ van sy een || lyn

or

]YB

/YZ||BC ; theoremprop.

of side one to||line[DY

DB

ZY

ZP :DYZIn

Z||C ; stellingewer.

ofΔ van sy een || lyn

or

DX

DA

YZ

BC

]

/parmaofsides [opp.BCZP

ZY

ZP

DX

DA

m|| van syeteenoorst

SR

S

S S / R

(5)

[11]

D

A

X Z

C B

Y P

10612

15

QUESTION / VRAAG 11

11.1 In NSR and NPQ

],,[ΔNQP|||ΔNRS

]so

Δainof[sumQPNS

]

segment same thein [QR

]/[givenQNPSNR

s

1

s

21

Δvan/Δvanm

segm sirkeldies inomtr

gegee

ee

e

S

S / R

R (3)

11.2 In NQR and NPS

],,[ΔNPS|||ΔNQR

]so

/Δainof[sumSR

]

segment /samethein[PQ

Q]NPSN[RSNPQNR

s

2

s

11

Δvan/Δvanm

segm sirkeldies inomtr

ee

e

S S

R (3)

11.3

NQ.PRNR.QRNR.PQ

PS)NQ(SR

PS.NQNQ.SRQR.NP NR.PQ

NQ.SRNR.PQ

ΔNQP]|||ΔNRS[PQ

SR

NQ

NR

PS.NQQR.NP

ΔNPS]|||ΔNQR[NP

NQ

PS

QR

S / R

S / R

S S (4) [10]

TOTAL / TOTAAL [150]